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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 45 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 45 Maths Textbook Solution.

Answers (1)

Answer:\frac{dy}{dx}=\frac{1}{2\sqrt{1-x^{3}}}

Hint:

\begin{aligned} &\frac{d}{d x}(\text { constant })=0 \\ &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \end{aligned}

Given:

y=\tan ^{-1}\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right\}

Solution:

Put,

x=\cos 2\theta

So,

\begin{aligned} y &=\tan ^{-1}\left[\frac{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}\right] \\ &=\tan ^{-1}\left(\frac{\sqrt{2 \cos ^{2} \theta}-\sqrt{2 \sin ^{2} \theta}}{\sqrt{2 \cos ^{2} \theta}+\sqrt{2 \sin ^{2} \theta}}\right) \end{aligned}

Using

\begin{aligned} &1+\cos 2 \theta=2 \cos ^{2} \theta \\ &y=\tan ^{-1}\left(\frac{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}\right) \\ &y=\tan ^{-1}\left(\frac{\sqrt{2}(\cos \theta-\sin \theta)}{\sqrt{2}(\cos \theta+\sin \theta)}\right) \end{aligned}

Dividing numerator and denominator by \cos \theta

\begin{aligned} &y=\tan ^{-1}\left(\frac{\frac{\cos \theta}{\cos \theta}-\frac{\sin \theta}{\cos \theta}}{\frac{\cos \theta}{\cos \theta}+\frac{\sin \theta}{\cos \theta}}\right) \\ &y=\tan ^{-1}\left(\frac{\cos \theta-\sin \theta}{\frac{\cos \theta}{\cos \theta+\sin \theta}} \cos \theta\right) \end{aligned}

Using

\begin{aligned} &\tan \theta=\frac{\sin \theta}{\cos \theta} \\ &y=\tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right) \\ &y=\tan ^{-1}\left(\frac{\tan \frac{n}{4}-\tan \theta}{1+\tan \frac{\pi}{4} \tan \theta}\right) \\ &y=\tan ^{-1}\left(\tan \left(\frac{n}{4}-\theta\right)\right) \\ &\text { Using } \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B} \end{aligned}

y=\frac{\pi-0}{4}-\theta                                                                                                                                        \left [ Using\: \tan ^{-1}\left ( \tan \theta \right ) =\theta \: if\: \theta \varepsilon \left [ -\frac{\pi}{2},\frac{\pi}{2} \right ]\right ]

y=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\text { Using } x=\cos 2 \theta]

Differentiating it with respect to x

\frac{\mathrm{dy}}{\mathrm{d} x}=0-\frac{1}{2}\left(\frac{-1}{\sqrt{1-\mathrm{x}^{2}}}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left\{\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}\left(\cos ^{-1} \mathrm{x}\right)=\frac{-1}{\sqrt{1-\mathrm{x}}^{2}}\right\}

\begin{aligned} &\frac{d y}{d x}=\frac{1}{2 \sqrt{1-x^{2}}} \\ &\frac{d y}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \end{aligned}

 

 

Posted by

infoexpert21

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